Dr. Alain Brizard
College Physics II (PY 211)
Gauss’s Law and Electric Potential
Textbook Reference: Chapter 22 – sections 1-3 & Chapter 23 – sections 1-8.
• Electric Flux
When a closed surface (denoted ∂V ) encloses an arbitrary charge distribution, the
surface integral of the electric field produced by the charge distribution defines the electric
flux
I
ΦE =
E · dA,
∂V
through the closed surface ∂V ; here, the vector dA denotes the infinitesimal surface element
(see Figure below).
Note that the vector dA is perpendicular to the surface ∂V and always points away from
the enclosed volume V . In some sense, the electric flux measures the net number of electric
field lines leaving volume V , i.e., it is equal to the number of outgoing field lines minus the
number of incoming field lines. In the Figure above, the number of outgoing field lines is
equal to the number of incoming field lines and, therefore, the electric flux for this example
is zero.
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• Gauss’s Law
According to Gauss’s Law, the electric flux through a closed surface ∂V is proportional
to the charge QV enclosed by the volume V :
I
QV
,
0
E · dA =
∂V
where 0 denotes the permittivity of free space. This formula can often be used to obtain
an expression for the electric field when the charge distribution has symmetry properties.
For example, a single charge q produces an electric field with spherical symmetry. Hence,
by spherical symmetry, we may assume that the electric field can be written as E = E(r) rb,
where the unit vector rb points in the direction of increasing radius. By choosing a Gaussian
surface ∂V defined as the surface of a sphere of radius r, with dA = rb (r2 sin θ dθ dϕ),
Gauss’s Law yields
I
E · dA = r2 E(r)
Z
π
sin θ dθ
0
∂V
Z
2π
dϕ
= 4π r2 E(r) =
0
q
,
0
where QV = q is the charge enclosed by the sphere of radius r. Hence, Gauss’s Law shows
that the electric field produced by a single charge q (located at the origin) is
q rb
,
4π0 r2
E =
as stated by Coulomb’s Law.
As another application of Gauss’s Law, we can now easily show that the electric field
inside a spherical shell of radius R carrying a uniform charge q is zero (since QV = 0
everywhere inside the shell), while the electric field outside the shell is everywhere identical
to the electric field due to a single charge q located at the center of the shell, i.e.,
E =
rb



4π r2 

0
for (r < R)
q/0 for (r > R)
If the charge q is distributed uniformly inside a sphere of radius R, on the other hand,
Gauss’s Law yields

 q (r/R)3 for (r < R)
rb 
E =
4π r2 

q/0
for (r > R)
where QV = q (r/R)3 is the charge enclosed by the Gaussian sphere of radius r. Hence, the
electric field inside a uniformly charged sphere of radius R is
Ein (r) =
2
qr
.
4π0 R3
We, therefore, conclude that, according to Gauss’s Law, the electric field outside a sphericallysymmetric charge distribution is independent of how the charge is distributed.
Additional applications of Gauss’s Law yield (a) for a cylindrically-symmetric infinitelylong uniform line of charge (with linear charge density λ):
E(r) (2π r `) =
λ`
0
→
E(r) =
λ
,
2π0 r
where the Gaussian surface was chosen as a cylinder of radius r and length `; and (b) for
an infinite charged plane (with surface charge density σ):
σA
σ
→ E =
.
0
2 0
Hence, for points sufficiently far away from the edges of a finite charged plane, the electric
field is uniform. Moreover, we can use this result to show that the electric field near any
conducting surface is
σ
b,
E =
n
0
b is a unit vector normal to the surface.
where n
E+ · A+ + E− · A− = 2 E A =
• Electric Potential
We have stated earlier that the electric field E = − ∇V can be expressed as the gradient
of the electric potential V . On the other hand, the electric potential V (r) at an arbitrary
point r can be calculated as the path integral involving the electric field E:
V (r) = −
Z
r
E(x) · dx,
r0
where r0 denotes a point where the electric potential is known to vanish. Consequently,
the potential difference Vb − Va between two points a and b is expressed in terms of the
electric field as
Z b
E(x) · dx.
Vb − Va = −
a
Hence, knowing that the electric field due to a single charge q located at the origin is
E(x) = kq x/|x|3 and assuming that the electric potential vanishes as r → ∞, we find the
electric potential at a distance r given as
V (r) = −
Z
r
∞
q
q ds
.
=
2
4π0 s
4π0 r
Next, the work Wab due to the electric force F = − q ∇V in carrying a charge q from
point a to point b is expressed in terms of the difference of the electric potential between
points a and b:
Wab =
Z
b
F · dx = − q
a
Z
b
a
3
∇V · dx = − q (Vb − Va ).
Hence, the electric force is a conservative force, since the work due to the electric force is
independent of the path taken, and we may define the electric potential energy of a charge
q located at point r as U (r) = q V (r). We further note that the electric potential at r
Wr→∞
q
V (r) =
is equal to the work done by the electric force per unit charge (which defines the electric
field) in bringing charge q from point r to infinity.
• Electric Potential due to an Arbitrary Charge Distribution
The electric potential at a field point r due to an arbitrary charge distribution is expressed as
Z
dq(r0)
V (r) =
,
r0 4π0 |r − r0 |
from which we easily recover Coulomb’s expression
E(r) = − ∇V (r) =
Z
r0
dq(r0 )
4π0
r − r0
|r − r0 |3
!
for the electric field. Lastly, we note that, although the calculation of the electric potential
by integration may seem to be simpler than that of the electric field (i.e., a scalar expression
is integrated instead of a vector expression), the electric potential is rarely calculated by
direct integration (other more powerful methods covered in advanced courses exist but are
beyond the scope of the present course).
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